DESIGN OF MAXIMUM ENERGY RECOVERY NETWORKS

DESIGN OF MAXIMUM ENERGY RECOVERY

NETWORKS

MER NETWORKS

• Networks featuring minimum utility usage are called MAXIMUM ENERGY RECOVERY (MER) Networks.

DIVISION AT THE PINCH

RECALL THAT• No heat is transferred through the pinch. • This makes the region above the pinch a

HEAT SINK region and the region below the pinch a HEAT SOURCE region.

Pinch

Minimum heating utility

Minimum cooling utilityHeat Source

Heat is released to cooling utility

Heat SinkHeat is obtained from the heating utility

7.5

12.0

14.0

0.0

-6.0

1.0

-4.0

4.0

3.0

9.0

-2.0

-2.0

14.0

1.5

10.0

T

H

CONCLUSION

• One can analyze the two systems separately, that is,

• Heat exchangers will not contain heat transfer across the pinch.

PINCH MATCHES• Consider two streams above the pinch

Q/FCpH > Q/FCpC

Tp- ∆TminTC,outQ

TH,inTp

TC,out= Tp- ∆Tmin + Q/FCpC

TH,in= Tp+Q/FCpH

But TH,in> TC,out+ ∆Tmin.

Thus replacing one obtains

FCpH < FCpC Golden rule for pinch matches above the pinch.

∆Tmin Violation when FCpH > FCpC∆Tmin

EXERCISE• A similar rule is obtained below the pinch. Derive it

and illustrate it.

ANSWER

TH,out

Tp- ∆Tmin

TC,inQ

Tp

TC,in= Tp- ∆Tmin - Q/FCpC

TH,out= Tp-Q/FCpH

But TH,out> TC,in + ∆Tmin.

Thus replacing one obtains

FCpC < FCpH Golden rule for pinch matches below the pinch.

∆Tmin∆Tmin

Violation when FCpC > FCpH

CONCLUSION

• Since matches at the pinch need to satisfy these rules, one should start locating these matches first. Thus, our first design rule:

START BY MAKING PINCH MATCHES

QUESTION

• Once a match has been selected how much heat should be exchanged?

ANSWER • As much as possible!• This means that one of the streams has its duty

satisfied!!

THIS IS CALLLED THE

TICK-OFF RULE

HANDS ON EXERCISE

Stream Type Supply T Target T ∆H F*Cp (oC) (oC) (MW) (MW oC-1)

Reactor 1 feed Cold 20 180 32.0 0.2

Reactor 1 product Hot 250 40 -31.5 0.15



T=140 0C

T=20 0C

∆H=27 MW

REACTOR 2

∆H=-30 MW

∆H=32 MW

REACTOR 1

∆H=-31.5 MW

T=230 0C

T=180 0C T=250 0C T=40 0C

T=200 0CT=80 0C

∆Tmin=10 oC PINCH=150 oC

HANDS ON EXERCISE

Stream Type Supply T Target T ∆H F*Cp (oC) (oC) (MW) (MW oC-1)





∆Tmin=10 oC PINCH=150 oC

200 0C

20 0C

40 0C

180 0C

250 0CH1

H2

C2

C1

230 0C

FCp=0.15

FCp=0.25

FCp=0.2

FCp=0.3

140 0C

80 0C

150 0C

ABOVE THE PINCH

• Which matches are possible?

200 0C

180 0C

250 0CH1

H2

C2

C1

230 0C

FCp=0.15

FCp=0.25

FCp=0.2

FCp=0.3140 0C

150 0C

ANSWER (above the pinch)

200 0C

180 0C

250 0CH1

H2

C2

C1

230 0C

FCp=0.15

FCp=0.25

FCp=0.2

FCp=0.3140 0C

150 0C

• The rule is that FCpH < FCpC . We therefore can only make the match H1-C1 and H2-C2.

ANSWER (above the pinch)

• The tick-off rule says that a maximum of 8 MW is exchanged in the match H1-C1 and as a result stream C1 reaches its target temperature.

• Similarly 12.5 MW are exchanged in the other match and the stream H2 reaches the pinch temperature.

200 0C

180 0C

203.3 0CH1

H2

C2

C1 181.7 0C

FCp=0.15

FCp=0.25

FCp=0.2

FCp=0.3140 0C

150 0C

8

12.5

230 0C

250 0C

BELOW THE PINCH

• Which matches are possible?

20 0C

40 0CH1

H2

C1

FCp=0.15

FCp=0.25

FCp=0.2 140 0C

80 0C

150 0C

The rule is that FCpC < FCpH . Thus, we can only make the match H2-C1

ANSWER (below the pinch)

• The tick-off rule says that a maximum of 17.5 MW is exchanged in the match H2-C1 and as a result stream H2 reaches its target temperature.

52.5 0C

40 0CH1

H2

C1

FCp=0.15

FCp=0.25

FCp=0.2 140 0C

80 0C

150 0C

17.5

20 0C

COMPLETE NETWORK AFTER PINCH MATCHES

• Streams with unfulfilled targets are colored.

200 0C

180 0C

203.3 0CH1

H2

C2

181.7 0C

FCp=0.15

FCp=0.25

FCp=0.2

FCp=0.3140 0C

8

12.5

52.5 0C C1

140 0C

80 0C

150 0C

17.5

250 0C

230 0C

20 0C

40 0C

WHAT TO DO NEXT?

• Away from the pinch, there is more flexibility to make matches, so the inequalities do not have to hold.

• The pinch design method leaves you now on your own!!!!!• Therefore, use your judgment as of what matches to select!!

200 0C

180 0C

203.3 0CH1

H2

C2

181.7 0C

FCp=0.15

FCp=0.25

FCp=0.2

FCp=0.3140 0C

8

12.5

52.5 0C C1

140 0C

80 0C

150 0C

17.5

250 0C

230 0C

20 0C

40 0C

ANSWER

• We first note that we will use heating above the pinch. Thus all hot streams need to reach their inlet temperature. We are then forced to look for a match for H1. Please locate it.

200 0C

180 0C

203.3 0CH1

H2

C2

181.7 0C

FCp=0.15

FCp=0.25

FCp=0.2

FCp=0.3140 0C

8

12.5

52.5 0C C1

140 0C

80 0C

150 0C

17.5

250 0C

230 0C

20 0C

40 0C

ANSWER

140 0C

• The match is H1-C1. We finally put a heater on the cold stream

200 0C

180 0C

H1

H2

C2

230 0C

FCp=0.15

FCp=0.25

FCp=0.2

FCp=0.38

12.5

52.5 0C C1

140 0C

80 0C

150 0C

17.5

77.5

H

250 0C

20 0C

40 0C

ANSWER • Below the pinch we try to have the cold streams start at

their inlet temperatures and we later locate coolers (one in this case).

140 0C

200 0C

180 0C

H1

H2

C2 230 0C

FCp=0.15

FCp=0.25

FCp=0.2

FCp=0.38

12.5

20 0C C1 140 0C

80 0C

150 0C

17.5

77.5

H6.5

C

10

40 0C250 0C

DISCUSSION

• State what are your objections to the pinch design method, so far!!

THESE ARE MINE

• It seems to imply building expertise• It does not seem to guarantee that one will always

obtain a good result. • What if there are unequal number of streams at the

pinch? • What if the inequalities are not satisfied at the

pinch?

GENERAL FORMULA FOR UNIT TARGETING

Nmin= (S-P)above pinch+ (S-P)below pinch

If we do not consider two separate problems, above and below the pinch we can get misleading results.

EXAMPLE

Nmin= (S-P)above pinch+ (S-P)below pinch ==(5-1) + (4-1) = 7

If we do not consider two separate problems Nmin= (6-1)= 5, which is wrong

Note: A heat exchanger network with 5 exchangers exists, but it is impractical and costly. This is beyond the scope of this course.

140 0C

200 0C

180 0C

H1

H2

C2

230 0C

FCp=0.15

FCp=0.25

FCp=0.2

FCp=0.38

12.5

20 0C C1 140 0C

80 0C

150 0C

17.5

77.5

H6.5

C

10

250 0C 40 0C

UNEQUAL NUMBER OF STREAMS AT THE PINCH

Above the pinch, we notice the following rule

SH ≤SC


Indeed, if the number of hot streams is larger than the number of cold streams, then no pinch matches are possible.

C2

C1

H1

H2

H3

Assume the matches H1-C1 and the matches H2-C2 have been selected. Since H3 needs to go to the pinch temperature, there is no cold stream left to match, even if there is portions of C1 or C2 that are left for matching. Such matching would be infeasible.

150 OC 100 OCFCp=0.15

FCp=0.25

FCp=0.2

FCp=0.4

FCp=0.1

90 OC

7.5

140 OC

10

Target=170 OC 127.5 OC

122.5 OCTarget=140 OC

130 OC

What is then, the solution?

ANSWER Split cold stream until the inequality is satisfied.

Notice that different combinations of flowrates in the split satisfy the inequality.

C2

C1

H1

H2

H3

150 OC 100 OCFCp=0.15

FCp=0.25

FCp=0.2

FCp=0.25

FCp=0.1

90 OC

7.5

140 OC

10

Target=170 OC127.5 OC

130 OCTarget=140 OC

130 OC

FCp=0.15 110 OC

3


A similar rule can be discussed below the pinch, that is,

SH ≥SC

and similar splitting of hot streams may be necessary.

INEQUALITY NOT SATISFIED

Consider the following case:

C2

C1

H1 150 OC 100 OC

FCp=0.5

FCp=0.2

FCp=0.4

90 OCTarget=170 OC

Target=140 OC

Suggest a solution.

ANSWER Split the hot stream

C2

C1

H1 150 OC 100 OCFCp=0.2

FCp=0.2

FCp=0.4

90 OC

15

Target=170 OC

Target=140 OC

FCp=0.3

10

140 OC

127.5 OC

SOLVE THE FOLLOWING PROBLEM

Below the Pinch :

C1

H2

H1 100 OC

FCp=0.5

FCp=0.3

FCp=0.7 90 OC

Target=40 OC

Target=20 OC

30 OC

ANSWER Below the Pinch :

30C1

H2

H1 100 OC

FCp=0.5

FCp=0.3

FCp=0.590 OC

Target=40 OC

Target=20 OC

30 OC

FCp=0.2

12

40 OC

60 OC

COMPLETE PROCEDUREABOVE THE PINCH

Start

SH≤SC

Split Cold Stream

FCpH≤FCpC

at pinch?

Place matches

Split Hot Stream

YesYes

NoNo

COMPLETE PROCEDUREBELOW THE PINCH

SH≥SC

Split Hot Stream

FCpH ≥ FCpCat pinch?

Split Cold Stream

Start

Place matches

YesYes

No No

HANDS ON EXERCISEType Supply T Target T F*Cp

(oC) (oC) (MW oC-1)

Hot 750 350 0.045

Hot 550 250 0.04

Cold 300 900 0.043

Cold 200 550 0.02

∆Tmin=50 oC

Minimum Heating Utility= 9.2 MW

Minimum Cooling Utility= 6.4 MW

ANSWER

750 0C

900 0C

H1

H2

C2

FCp=0.045

FCp=0.04

FCp=0.043

FCp=0.02

500 0C

8

C1

250 0C

61

9.2

300 0C

8.6

0.4

550 0C

550 0C

6

350 0C

200 0C

FCp=0.04

686.05 0C 400 0C

358.89 0C

ANSWER

We are minimizing the number of matches. We saw already that different answers can be obtained if separate regions are not considered.

DESIGN OF MAXIMUM ENERGY RECOVERY NETWORKS

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